Engineering Mathematics Miscellaneous


Engineering Mathematics Miscellaneous

Engineering Mathematics

  1. The differential equation
    dy
    + 4y = 5
    dx

    is valid in the domain 0 ≤ x ≤ 1 with y(0) = 2.25. The solution of the differential equation is









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    dy
    + 4y = 5
    dx

    I.F. = e∫4 dx = e4 x
    ⇒ y × I.F. = ∫ 5 × (I.F.)dx + C
    ⇒ y . e4 x = 5.
    e4 x
    + C
    4

    ⇒ 2.25 × 1 = 5.
    e0
    + C
    4

    ⇒ C = 1
    y . e4 x = 5.
    e4 x
    + 1
    4

    ⇒ y =
    5
    + e-4 x
    4

    ⇒ y = e-4 x + 1.25

    Correct Option: B

    dy
    + 4y = 5
    dx

    I.F. = e∫4 dx = e4 x
    ⇒ y × I.F. = ∫ 5 × (I.F.)dx + C
    ⇒ y . e4 x = 5.
    e4 x
    + C
    4

    ⇒ 2.25 × 1 = 5.
    e0
    + C
    4

    ⇒ C = 1
    y . e4 x = 5.
    e4 x
    + 1
    4

    ⇒ y =
    5
    + e-4 x
    4

    ⇒ y = e-4 x + 1.25


  1. For the equation
    dy
    + 7x2y = 0 , if y(0) = 3 / 7 ,
    dx

    then the value of y(1) is









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    From question, we have

    dy
    + 7x2y = 0 , y(0) =
    3
    , y(1) = ?
    dx7

    dy
    = -7x2y
    dx

    dy
    = -7x2dx
    y

    ln y = -
    7x3
    + C
    3

    y = e{ (-7x3 / 3) + C}
    y(0) = eC =
    3
    7

    C = ln
    3
    7

    ln y = -
    7x3
    + ln
    3
    37

    ln y - ln
    3
    = -
    7x3
    73

    ln
    7y
    = -
    7x3
    33

    y =
    3
    e (-7x3/ 3)
    7

    ∴ y(1) =
    3
    e (-7/ 3)
    7

    Correct Option: A

    From question, we have

    dy
    + 7x2y = 0 , y(0) =
    3
    , y(1) = ?
    dx7

    dy
    = -7x2y
    dx

    dy
    = -7x2dx
    y

    ln y = -
    7x3
    + C
    3

    y = e{ (-7x3 / 3) + C}
    y(0) = eC =
    3
    7

    C = ln
    3
    7

    ln y = -
    7x3
    + ln
    3
    37

    ln y - ln
    3
    = -
    7x3
    73

    ln
    7y
    = -
    7x3
    33

    y =
    3
    e (-7x3/ 3)
    7

    ∴ y(1) =
    3
    e (-7/ 3)
    7



  1. If y is the solution of the differential equation y3
    dy
    + x3 = 0 ,
    dx

    y(0) = 1 the value of y(–1) is









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    y3
    dy
    = -x3
    dx

    y3 dy = -x3 dx
    ∫ y3 dy = -∫ x3 dx
    y4
    =
    − x4
    + c
    44

    x4 + y4
    = c
    4

    y(0) = 1 ,
    0 + 1
    = c
    4

    c =
    1
    4

    x4 + y4 = 1
    y = 4√1 - x4
    when x = -1
    y = 0

    Correct Option: C

    y3
    dy
    = -x3
    dx

    y3 dy = -x3 dx
    ∫ y3 dy = -∫ x3 dx
    y4
    =
    − x4
    + c
    44

    x4 + y4
    = c
    4

    y(0) = 1 ,
    0 + 1
    = c
    4

    c =
    1
    4

    x4 + y4 = 1
    y = 4√1 - x4
    when x = -1
    y = 0


  1. Consider the following differential equation :
    dy
    = -5y ; initial condition; y = 2 at t = 0 The value of y at t = 3 is
    dt










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    dy
    = -5y
    dt

    dy
    = -5dt
    y

    By Integrating above equation, we get
    ln y = 5t + c
    at t = 0 , y = 2
    ln 2 = c
    So, ln y = – 5t + ln 2
    ln
    y
    = -5t
    2

    y
    = e-5t
    2

    y = 2e-5t
    at t = 3
    y = 2 e-15

    Correct Option: C

    dy
    = -5y
    dt

    dy
    = -5dt
    y

    By Integrating above equation, we get
    ln y = 5t + c
    at t = 0 , y = 2
    ln 2 = c
    So, ln y = – 5t + ln 2
    ln
    y
    = -5t
    2

    y
    = e-5t
    2

    y = 2e-5t
    at t = 3
    y = 2 e-15



  1. The distance between the origin and the point nearest it on the surface z2 = 1 + xy is









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    Take a point on the curve z = 1, x = 0, y = 0
    Length between origin and this point
    l = √( 1- 0) + (0 - 0) + (0 - 0 ) = 1
    This is minimum length because all options have length greater than l.

    Correct Option: A

    Take a point on the curve z = 1, x = 0, y = 0
    Length between origin and this point
    l = √( 1- 0) + (0 - 0) + (0 - 0 ) = 1
    This is minimum length because all options have length greater than l.